C++11中std::forward的使用
std::forward argument: Returns an rvalue reference to arg if arg is not an lvalue reference; If arg is an lvalue reference, the function returns arg without modifying its type.
std::forward:This is a helper function to allow perfect forwarding of arguments taken as rvalue references to deduced types, preserving any potential move semantics involved.
std::forward<T>(u)有两个参数:T 与 u。当T为左值引用类型时,u将被转换为T类型的左值,否则u将被转换为T类型右值。如此定义std::forward是为了在使用右值引用参数的函数模板中解决参数的完美转发问题。
std::move是无条件的转为右值引用,而std::forward是有条件的转为右值引用,更准确的说叫做Perfect forwarding(完美转发),而std::forward里面蕴含着的条件则是Reference Collapsing(引用折叠)。
std::move不move任何东西。std::forward也不转发任何东西。在运行时,他们什么都不做。不产生可执行代码,一个比特的代码也不产生。
std::move和std::forward只是执行转换的函数(确切的说应该是函数模板)。std::move无条件的将它的参数转换成一个右值,而std::forward当特定的条件满足时,才会执行它的转换。
std::move表现为无条件的右值转换,就其本身而已,它不会移动任何东西。 std::forward仅当参数被右值绑定时,才会把参数转换为右值。 std::move和std::forward在运行时不做任何事情。
下面是从其他文章中copy的测试代码,详细内容介绍可以参考对应的reference:
#include "forward.hpp" #include <utility> #include <iostream> #include <memory> #include <string> ////////////////////////////////////////////// // reference: http://en.cppreference.com/w/cpp/utility/forward struct A { A(int&& n) { std::cout << "rvalue overload, n=" << n << "\n"; } A(int& n) { std::cout << "lvalue overload, n=" << n << "\n"; } }; class B { public: template<class T1, class T2, class T3> B(T1&& t1, T2&& t2, T3&& t3) : a1_{ std::forward<T1>(t1) }, a2_{ std::forward<T2>(t2) }, a3_{ std::forward<T3>(t3) } { } private: A a1_, a2_, a3_; }; template<class T, class U> std::unique_ptr<T> make_unique1(U&& u) { return std::unique_ptr<T>(new T(std::forward<U>(u))); } template<class T, class... U> std::unique_ptr<T> make_unique(U&&... u) { return std::unique_ptr<T>(new T(std::forward<U>(u)...)); } int test_forward1() { auto p1 = make_unique1<A>(2); // rvalue int i = 1; auto p2 = make_unique1<A>(i); // lvalue std::cout << "B\n"; auto t = make_unique<B>(2, i, 3); return 0; } //////////////////////////////////////////////////////// // reference: http://www.cplusplus.com/reference/utility/forward/ // function with lvalue and rvalue reference overloads: void overloaded(const int& x) { std::cout << "[lvalue]"; } void overloaded(int&& x) { std::cout << "[rvalue]"; } // function template taking rvalue reference to deduced type: template <class T> void fn(T&& x) { overloaded(x); // always an lvalue overloaded(std::forward<T>(x)); // rvalue if argument is rvalue } int test_forward2() { int a; std::cout << "calling fn with lvalue: "; fn(a); std::cout << '\n'; std::cout << "calling fn with rvalue: "; fn(0); std::cout << '\n'; return 0; } ////////////////////////////////////////////////////// // reference: http://stackoverflow.com/questions/8526598/how-does-stdforward-work template<class T> struct some_struct{ T _v; template<class U> some_struct(U&& v) : _v(static_cast<U&&>(v)) {} // perfect forwarding here // std::forward is just syntactic sugar for this }; int test_forward3() { /* remember the reference collapsing rules(引用折叠规则): 前者代表接受类型,后者代表进入类型,=>表示引用折叠之后的类型,即最后被推导决断的类型 TR R T& &->T& // lvalue reference to cv TR -> lvalue reference to T T& &&->T& // rvalue reference to cv TR -> TR (lvalue reference to T) T&& &->T& // lvalue reference to cv TR -> lvalue reference to T T&& &&->T&& // rvalue reference to cv TR -> TR (rvalue reference to T) */ some_struct<int> s1(5); // in ctor: '5' is rvalue (int&&), so 'U' is deduced as 'int', giving 'int&&' // ctor after deduction: 'some_struct(int&& v)' ('U' == 'int') // with rvalue reference 'v' bound to rvalue '5' // now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int&&>(v)' // this just turns 'v' back into an rvalue // (named rvalue references, 'v' in this case, are lvalues) // huzzah, we forwarded an rvalue to the constructor of '_v'! // attention, real magic happens here int i = 5; some_struct<int> s2(i); // in ctor: 'i' is an lvalue ('int&'), so 'U' is deduced as 'int&', giving 'int& &&' // applying the reference collapsing rules yields 'int&' (& + && -> &) // ctor after deduction and collapsing: 'some_struct(int& v)' ('U' == 'int&') // with lvalue reference 'v' bound to lvalue 'i' // now we 'static_cast' 'v' to 'U&&', giving 'static_cast<int& &&>(v)' // after collapsing rules: 'static_cast<int&>(v)' // this is a no-op, 'v' is already 'int&' // huzzah, we forwarded an lvalue to the constructor of '_v'! return 0; } //////////////////////////////////////////////////// // reference: https://oopscenities.net/2014/02/01/c11-perfect-forwarding/ void sum(int a, int b) { std::cout << a + b << std::endl; } void concat(const std::string& a, const std::string& b) { std::cout<< a + b << std::endl; } void successor(int a, int& b) { b = ++a; } template <typename PROC, typename A, typename B> void invoke(PROC p, A&& a, B&& b) { p(std::forward<A>(a), std::forward<B>(b)); } int test_forward4() { invoke(sum, 10, 20); invoke(concat, "Hello", "world"); int s = 0; invoke(successor, 10, s); std::cout << s << std::endl; return 0; }
GitHub:https://github.com/fengbingchun/Messy_Test
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